Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Amongst the following, the linear species is

The quest to determine the molecular geometry of a species is one of the most fundamental and exciting challenges in chemistry. It’s like solving a microscopic puzzle where the rules are governed by the repulsion of electrons. In this problem, we are tasked with finding the linear species among four candidates: ${\text{NO}}_2$, ${\text{Cl}}_2\text{O}$, ${\text{O}}_3$, and ${\text{N}}_3^{-}$.

To crack this, we will rely on our trusty VSEPR (Valence Shell Electron Pair Repulsion) theory. The core idea is simple: electron pairs around a central atom want to stay as far away from each other as possible to minimize repulsion.

Steric Number

Before we dive into the molecules, let's recall the master formula for predicting geometry. We need to calculate the Steric Number (SN) for the central atom:

The steric number tells us the hybridization and the base geometry: $\text{SN}=2⟹sp\text{ hybridization}⟹\text{Linear}$ $\text{SN}=3⟹s{p}^2\text{ hybridization}⟹\text{Trigonal Planar}$ * $\text{SN}=4⟹s{p}^3\text{ hybridization}⟹\text{Tetrahedral}$

Now, let's put each molecule under the microscope!

1. Nitrogen Dioxide (${\text{NO}}_2$) Imagine the central nitrogen atom. It forms two $\sigma$ bonds with the oxygen atoms. But nitrogen has 5 valence electrons. After forming the bonds, it is left with one unpaired, odd electron.

Even though it's just a single electron, it occupies a hybrid orbital.

This gives an $s{p}^2$ hybridization. The repulsion from this odd electron pushes the two oxygen atoms down, resulting in a bent shape.

2. Dichlorine Monoxide (${\text{Cl}}_2\text{O}$) Here, the central atom is oxygen. Oxygen has 6 valence electrons. It uses 2 electrons to form $\sigma$ bonds with the chlorine atoms, leaving 4 electrons, which make up 2 lone pairs.

This corresponds to $s{p}^3$ hybridization. The two bulky lone pairs exert a strong repulsive force on the bonding pairs, squeezing them together and creating a bent shape.

3. Ozone (${\text{O}}_3$) In ozone, the central oxygen atom forms a double bond with one oxygen (1 $\sigma$ bond) and a single coordinate bond with the other (1 $\sigma$ bond). Out of its 6 valence electrons, it has 1 lone pair remaining.

With an $s{p}^2$ hybridization, the lone pair repels the bonding pairs, giving ozone a bent shape as well.

The Azide Ion

4. The Azide Ion (${\text{N}}_3^{-}$) Finally, let's look at the azide ion. The central nitrogen atom forms two double bonds with the terminal nitrogen atoms. Each double bond counts as one $\sigma$ bond.

What about lone pairs? The central nitrogen has a formal charge of $+1$ (it shares 4 electrons and "owns" 4, compared to its usual 5). This means it has zero lone pairs!

A steric number of 2 means the central nitrogen is $sp$ hybridized. With no lone pairs to push the bonds around, the two terminal nitrogen atoms stretch out as far as possible—exactly $180^{\circ }$ apart.

This results in a perfectly linear shape.

By systematically applying VSEPR theory and calculating the steric numbers, we have successfully unmasked the linear molecule. The azide ion (${\text{N}}_3^{-}$) stands out as the only species with a linear geometry among the given options.